Typescript-Algorithms
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    Variable binary_tree_inorder_traversalConst

    binary_tree_inorder_traversal: (root: null | TreeNode) => number[] = inorderTraversal

    94.二叉树的中序遍历

    给定一个二叉树的根节点 root ,返回 它的 中序 遍历 。


    示例1

    输入: root = [1,null,2,3]
    输出: [1,3,2]


    输入: root = []
    输出: []


    输入: root = [1]
    输出: [1]


    • 树中节点数目在范围 [0, 100]
    • -100 <= Node.val <= 100

    Type declaration

      • (root: null | TreeNode): number[]
      • Definition for a binary tree node. class TreeNode { val: number left: TreeNode | null right: TreeNode | null constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) { this.val = (val===undefined ? 0 : val) this.left = (left===undefined ? null : left) this.right = (right===undefined ? null : right) } }

        Parameters

        • root: null | TreeNode

        Returns number[]

    二叉树的中序遍历顺序是:左根右

    /**
    * Definition for a binary tree node.
    * class TreeNode {
    * val: number
    * left: TreeNode | null
    * right: TreeNode | null
    * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    * this.val = (val===undefined ? 0 : val)
    * this.left = (left===undefined ? null : left)
    * this.right = (right===undefined ? null : right)
    * }
    * }
    */

    function inorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];

    function dfs(node: TreeNode | null) {
    if (!node) {
    return;
    }

    dfs(node.left);
    ans.push(node.val);
    dfs(node.right);
    }

    dfs(root);

    return ans;
    };